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EXAMPLE. The fixed axis, a b, of the prolate spheroid a c b d, is 32 inches, and the revolving axis, cd, is 20 inches; required the solid contents.

202 X 32 X.5236 : = 6702.08 cubic inches =
6702.0817283.88- cubic feet.

Ans.

To find the solidity of the segment of a spheroid.

RULE.When the base of the segment is parallel to the shorter axis of the spheroid. From three times the length of the longer axis, subtract twice the height of the segment, and multiply the difference by the square of the height, multiplied by the square of the shorter axis, multiplied by .5236, and divide the product by the square of the longer axis; the quotient will be the solid contents of the segment.

RULE. When the base of the segment is parallel to the longer axis of the spheroid. — From three times the length of the shorter axis subtract twice the height of the segment, and multiply the difference by the square of the height multiplied by the longer axis, multiplied by .5236, and divide the product by the shorter axis; the quotient will be the solidity.

EXAMPLE. -The longer axis, a b, of the spheroid a c b d, being 32 inches, and the shorter, c d, 20 inches, what is the solidity of the segment ef a, whose base, eƒ, is 12 inches, and height, ga, 4 inches? 32 X 3-8=88 X 16 X 202 X.5236294891.52 1024 = 287.98

cub. inches. Ans.

To find the solidity of the middle frustum of a spheroid. RULE. When the ends of the frustum are parallel to the revolving To the square of the diameter of either end add twice the

axis.

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square of the revolving axis, and multiply

the sum by the length of the frustum multi

plied by .2618; the product will be the solidity; or the capacity, the frustum being a cask, and the measures taken of the interior.

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the revolving axis, c d, is 23 inches, and the length of the frustum, a b, is 20 inches; required the cubic capacity of the frustum. 2322182 1382 X 20 X .26187236.152 1728 4.187 cubic feet. Ans.

RULE.

=

OF SPINDLES AND CONOIDS.

To find the solidity of an elliptic spindle.

-To the square of twice the diameter of the spindle, at its length, add the square of its greatest diameter, and multiply the sum by the length, multiplied by .1309; the product will be the solidity.

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The greatest diameter,

f

cd, of the elliptic spindle a cb d, is
12 inches, the diameter at its length,
ef, is 8 inches, and the length of the
spindle, a b, is 40 inches; required its solidity.

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17, and 172 + 12 2 X 40 X .1309 = 2267.188 cubic. in.

Ans.

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c d, at the middle of the parabolic spindle Ed F c, is 15

E

a

d

F

inches, and the length of the spindle, E F, is 40 inches; required its solidity.

152 X 40 X .41888 3769.92 cubic inches. Ans.

To find the solidity of the middle frustum of a parabolic spindle. RULE. To eight times the square of the greatest diameter, add three times the square of the least diameter, and four times the product of the two diameters, and multiply the sum by the length of the frustum, multiplied by .05236, and the product is the solidity.

EXAMPLE. The greatest diameter, c d, of the frustum a cb d, is 28 inches, the least diameter (that of either end) is 20 inches, and the length of the frustum, a b, is 40 inches; required the solidity. 282 × 8+202 × 3 + 28 × 20 X 4 X 40 X .05236

cubic feet. Ans.

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1728

To find the solidity of a paraboloid, or parabolic conoid.

RULE. - Multiply the area of the base by half the altitude; or, multiply the square of the diameter of the base by the altitude multiplied by .3927, and the product will be the solidity.

EXAMPLE. The diameter of the base, cd, of the paraboloid, c d g, is 40 inches, and its height, fg, is 40 inches; required its solidity.

a

402 X .7854 X 2025132.8 cubic inches. Ans.

To find the solidity of a frustum of a paraboloid.

RULE. Add the squares of the diameters of the two bases together, and multiply the sum by the distance between the bases, multiplied by .3927; the product will be the solidity.

EXAMPLE. The diameter of the base, c d, of the frustum, c d ba, is 40 inches, the diameter of the base, a b, is 22 inches, and the height of the frustum, f e, is 26 inches; required the solidity, or capacity. 402222 2084 X 26 X .3927=21278 cubic inches. Ans.

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To find the solidity of a hyperboloid or hyperbolic conoid. RULE. To the square of half the diameter of the base add the square of the diameter midway between the base and vertex, and multiply the sum by the distance between the base and vertex, multiplied by .5236; the product will be the solidity.

EXAMPLE. The diameter of the base, a b, of the hyperboloid, a bf, is

a

40 inches, the diameter midway between the base and vertex is 26 inches, and the altitude of the figure, e ƒ, is 24 inches; required its solidity.

202 + 262 = 1076 × 24 × .5236 = 13521.4+ cubic in. Ans.

To find the solidity of a frustum of a hyperboloid.

RULE. To the square of half the diameter of one end, add the square of half the diameter of the other end, and the square of the diameter midway between the two ends, and multiply the sum by the length of the frustum, multiplied by .5236; the product is the solidity.

EXAMPLE. The diameter of the base, a b, of the frustum, a b d c, is 40 inches, the diameter of the base, c d, is 17 inches, the diameter at the point equally distant from either base, is 29.1 inches, and the length of the frustum, e r, is 18 inches; required the solidity.

202+8.52+29.12:

NOTE.

1319 X 18 X .5236 12432 cub. in. Ans.

The solidity or capacity of a frustum of any of the conic sections may be found by first finding the mean diameter, which, on being squared and multiplied by the length of the frustum decreased by being multiplied by .7854, gives the cubic contents. Instance, the last example:

401723 X .55* = 12.65 +17=29.65

mean diameter, and 29.652 X 18 X .7854 12428 cub. in. Ans.

To find the surface of a cylindrical ring. RULE.-To the inner diameter of the ring add its thickness, and multiply the sum by the thickness, multiplied by 9.8696; the product is the surface.

EXAMPLE. The inner diameter, bc, of a cylindrical ring is 12 inches, and its thickness, ab, is three inches; required the superfices.

D

123 15 X 3X 9.8696

444 sq. inches. Ans.

RULE.

To find the solidity of a cylindrical ring.

To the inner diameter of the ring add its thickness, and multiply the sum by the square of the thickness, multiplied by 2.4674; the product is the solidity.

EXAMPLE. The inner diameter, b c, is 12 inches, and the thick. ness, a b, is 3 inches'; the solidity is required.

12 + 3 × 32 × 2.4674

= 333.1 cubic inches. Ans.

*For the co efficient multipliers, and rule in detail, see GAUGING, Rule 5,

PROMISCUOUS EXAMPLES IN GEOMETRY.

A brace in a building is to have a run of 3 feet on the post and 4 feet on the plate; what must be the length of the brace, and at what bevels must the shoulders of the tenons be cut?

√(42+32) = 5 feet, length of brace. Ans.

3 × 86* ÷ (5+) = 36°.86, bevel for end at plate.† Ans. 53°.14, bevel for end at post. Ans.

90° 36°.86

=

NOTE.-To convert the decimal of a degree into minutes, multiply it by 60.

The roof of a building 30 feet wide is to have a third pitch; what must be the length of the rafters, supposing the sides of the roof are to be equal, and at what bevels must the ends of the rafters be cut?

30215 feet, half width of building.

30 ÷ 3 = 10 feet, height of roof; then

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56°.3 X 2 1120.6 90°

=

56°.3, bevel for

22°.6, bevel for top or peak,

(direct with that at the foot,) if one rafter is to be let into the other; or,

22.6 ÷ 2 = 11°.3, bevel for top, if the ends are to be halved together.

A building 26 feet wide, has a roof 7 feet high, and the ridge is vertical to a line 10 feet from one side; what is the length of the rafters?

26

1016, and (162 + 72) = 17.46 feet, length of each rafter in longest side of roof.

✔(10272) = 12.2 feet, length of each rafter in shortest side

of roof.

Two perpendicular walls are standing on a plane, 80 feet apart; one is 20 feet high, and the other 30; at what distance on the plane,

*See TRIGONOMETRY, p. 204.

† 900 is the 0 or starting point of a bevel; the nearer an angle is, therefore, to a rightangle the less the bevel.

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